Interpret the circuit

Since the question says two sources of equal emf are connected to an external resistance and then refers to the terminal potential difference of one source being zero, the standard interpretation is that the two cells are connected in parallel across the external resistance $R$.

Let each source have emf $E$. Their internal resistances are $R_1$ and $R_2$, with $R_2>R_1$.

Condition for zero terminal potential difference across source with internal resistance $R_2$

For a cell delivering current, terminal voltage is $V = E - Ir.$

For the source with internal resistance $R_2$, terminal voltage is zero: $0 = E - I_2R_2$ so $I_2 = \frac{E}{R_2}.$

This means the entire emf of that source is dropped inside its internal resistance.

Common terminal voltage in parallel connection

Since the two sources are in parallel, the terminal voltage across both sources and the external resistor is the same.

Given that the terminal voltage of the $R_2$ source is zero, the voltage across the external resistor is also $V=0.$

Hence current through external resistance is $I_R = \frac{V}{R}=0.$

Apply current law at the junction

Since no current flows through the external resistor, the current supplied by one source must be balanced by the current absorbed by the other source.

Let current in the source with internal resistance $R_1$ be $I_1$.

For that source also terminal voltage is zero, so $0 = E - I_1R_1$ giving $I_1 = \frac{E}{R_1}.$

But then both sources would attempt to send current into the same zero-voltage terminals, which is inconsistent unless we analyze the equivalent condition more carefully.

Use equivalent source method

Two equal-emf cells $E$ with internal resistances $R_1$ and $R_2$ in parallel are equivalent to a single source of emf $E$ with internal resistance $r_{eq} = \frac{R_1R_2}{R_1+R_2}.$

Current through external resistor is $I = \frac{E}{R+r_{eq}}.$

Terminal voltage across the combination is $V = IR = \frac{ER}{R+r_{eq}}.$

This terminal voltage is also the terminal voltage of each cell.

For the source with internal resistance $R_2$, current through that branch is $I_2 = \frac{E-V}{R_2}.$

Given terminal voltage across that source is zero: $V=0.$

But this would force $R=0$, which is not among the options. So the intended meaning must be different.

Correct interpretation: cells are connected in series aiding

The options strongly suggest the intended arrangement is two equal emf sources in series with external resistance $R$.

Then total emf is $2E$ and total internal resistance is $R_1+R_2.$

Circuit current is $I = \frac{2E}{R+R_1+R_2}.$

Use zero terminal voltage condition for the source with internal resistance $R_2$

Its terminal voltage is $V_2 = E - IR_2.$

Given $V_2=0$, $E - IR_2 = 0$ $I = \frac{E}{R_2}.$

Equate the two expressions for current

$\frac{2E}{R+R_1+R_2} = \frac{E}{R_2}.$

Cancel $E$: $\frac{2}{R+R_1+R_2} = \frac{1}{R_2}.$

Cross-multiply: $2R_2 = R+R_1+R_2$ $R = R_2 - R_1.$

Check options

This matches: $\boxed{R = R_2 - R_1}$

So the correct option is A.